422 XV NORMED ALGEBRAS AND SPECTRAL THEORY (1) C is an eigenvalue of T (the set of these is called the point- spectrum of T) ; (2) f is not an eigenvalue of T (which implies that T £/ is a bijective mapping of dom(T) onto a subspace L of E), the subspace L is dense in E, but (jT C/)"1 is not continuous on L (the set of these values is called the con- tinuous spectrum of T) ; (3) C is not an eigenvalue of T and the subspace L is not dense in E (the set of these values is called the residual spectrum of T). For example, for the operator defined in (11.1.1), the point-spectrum is empty, the continuous spectrum is the unit circle U : |C| = 1, and the residual spectrum is the open disk |C| < 1 (Section 11.1, Problem 4). (15.12.11) The spectrum of a closed operator T on E is closed in C, and the mapping C^^r(0 ofC - Sp(T) into £(E) is analytic. Let Co e C Sp(T), and put a = ||jRT(Co)ll- Then the continuous operator = (/- (C - Co)JĞr(Co))"%(Co)isdefinedforallCsatisfying|C--C0|<fl-l> and is analytic in this disk (83.2.1). For each x e dom(T), we have *T(Co)(^ - C/) ' * = * + (Co - and therefore V(Q(T - U) - x = x. For each integer n^Q, let so that K(C) is the limit of the sequence (Fn(Q) in Jz?(E). By definition, we have Fn(Q y e dom(r) for all y e E and all n > 0, and (T - CTOC) -y = (C - C0)K,(0 y + (T - C hence (r CO^AO ' J tends to y as n -ğ +00. Since by hypothesis 7 is closed, this implies that F(0 y e dom(T) and (T - C/) P(0 y = y- This shows that T C/ is a bijective linear mapping of dom(T) onto E and that the inverse mapping is K(0. Hence C # Sp(T) and F(Q = ^r(0> which completes the proof. (15.12.12) £#£ N be a (not necessarily bounded) normal operator on E. With the notation of (15. 12.8), we have Sp(N) = (J Sp(7V,,); the point-spectrum ofN n is the union of the point-spectra of the Nn , and the residual spectrum of N is empty. For N to be self-adjoint, it is necessary and sufficient that its spectrum should be real.d if the inverse linear